Question

You push a wagon of mass M up a hill of angle ?. The distance you push the wagon up the hill is d. This is the distance up the slope, not the vertical height the wagon moves. You push with a constant force equal to 1/3 the force of gravity on the wagon directly up the hill. The wagon starts at rest at the bottom of the hill. There is a small frictional force between the hill and the wagon wheels that you can model with a coefficient of friction LaTeX: \mu ?. This acts just like kinetic friction. M = 38.2 kg ? =3.1o d = 7.80 m FPush = FG/3 LaTeX: \mu ? = 0.067 Assume air drag on the wagon can be neglected in this problem. What is the kinetic energy of the wagon at the top of the hill? Give your answer in Joules to at least three significant digits to avoid being counted off due to rounding. Do not include units in your answer.

Answer 1

Solution:

Mass = M

Force of gravity = F_G .= Mg

Applied force = F = F_G /3 = Mg /3

Coefficient of friction =

The component of force of gravity down the hill = F_G sin _{= Mg sin}

Friction force =    N where N = F_G cos = Mg cos is the Normal force.

Net force on the object F net = Ma

Mg /3 - Mg sin - Mg cos = Ma

=> a = g/3 - g sin - g cos = 9.81 (1/3 - sin 3.1 - (0.067) cos 3.1) = 0.212 m/s^2

Distance moved up along the incline = L=d= 7.8m

FInal velocity is calculated using vf^2 =vo^2 +2ad

vf^2 = 0^2+ 2*0.212 * 7.8 => final velocity =vf = 1.82 m/s

Kinetic energy = 1/2 mv^2 = 1/2 * 38.2 * (1.82)^2 = 63.16752

= 63.2 joules (3 significant digits)